CUET Preparation Today
The area of the region bounded by the lines $\frac{x}{7 \sqrt{3} a}+\frac{y}{b}=4, x=0$ and $y=0$ is: |
$56 \sqrt{3} ab$ $56 a$ $\frac{a b}{2}$ $3 ab$ |
$56 \sqrt{3} ab$ |
The correct answer is Option (1) → $56 \sqrt{3} ab$ $\frac{x}{7 \sqrt{3} a}+\frac{y}{b}=4$
area = $\frac{1}{2}×28\sqrt{3}a×4b$ $=56 \sqrt{3} ab$ |
